我希望我将问题简化为一个简单且可重复的测试用例。源(here)包含10个相同的简单循环副本。每个循环的形式如下:
#define COUNT (1000 * 1000 * 1000)
volatile uint64_t counter = 0;
void loopN(void) {
for (int j = COUNT; j != 0; j--) {
uint64_t val = counter;
val = val + 1;
counter = val;
}
return;
}
' volatile'变量很重要,因为它强制值在每次迭代时从内存中读取和写入。每个循环使用' -falign-loops = 64'对齐到64个字节。并生成相同的程序集,但与全局的相对偏移量除外:
400880: 48 8b 15 c1 07 20 00 mov 0x2007c1(%rip),%rdx # 601048 <counter>
400887: 48 83 c2 01 add $0x1,%rdx
40088b: 83 e8 01 sub $0x1,%eax
40088e: 48 89 15 b3 07 20 00 mov %rdx,0x2007b3(%rip) # 601048 <counter>
400895: 75 e9 jne 400880 <loop8+0x20>
我在Intel Haswell i7-4470上运行Linux 3.11。我用GCC 4.8.1和命令行编译程序:
gcc -std=gnu99 -O3 -falign-loops=64 -Wall -Wextra same-function.c -o same-function
我还在源代码中使用属性((noinline))来使程序集更加清晰,但这不是观察问题所必需的。我找到了shell循环中最快和最慢的函数:
for n in 0 1 2 3 4 5 6 7 8 9;
do echo same-function ${n}:;
/usr/bin/time -f "%e seconds" same-function ${n};
/usr/bin/time -f "%e seconds" same-function ${n};
/usr/bin/time -f "%e seconds" same-function ${n};
done
它产生的结果与运行一致约为1%,最快和最慢函数的确切数量根据确切的二进制布局而变化:
same-function 0:
2.08 seconds
2.04 seconds
2.06 seconds
same-function 1:
2.12 seconds
2.12 seconds
2.12 seconds
same-function 2:
2.10 seconds
2.14 seconds
2.11 seconds
same-function 3:
2.04 seconds
2.04 seconds
2.05 seconds
same-function 4:
2.05 seconds
2.00 seconds
2.03 seconds
same-function 5:
2.07 seconds
2.07 seconds
1.98 seconds
same-function 6:
1.83 seconds
1.83 seconds
1.83 seconds
same-function 7:
1.95 seconds
1.98 seconds
1.95 seconds
same-function 8:
1.86 seconds
1.88 seconds
1.86 seconds
same-function 9:
2.04 seconds
2.04 seconds
2.02 seconds
在这种情况下,我们看到loop2()是最慢的执行之一,而loop6()是最快的之一,差异只有10%以上。我们通过使用不同的方法重复测试这两种情况来重新确认这一点:
nate@haswell$ N=2; for i in {1..10}; do perf stat same-function $N 2>&1 | grep GHz; done
7,180,104,866 cycles # 3.391 GHz
7,169,930,711 cycles # 3.391 GHz
7,150,190,394 cycles # 3.391 GHz
7,188,959,096 cycles # 3.391 GHz
7,177,272,608 cycles # 3.391 GHz
7,093,246,955 cycles # 3.391 GHz
7,210,636,865 cycles # 3.391 GHz
7,239,838,211 cycles # 3.391 GHz
7,172,716,779 cycles # 3.391 GHz
7,223,252,964 cycles # 3.391 GHz
nate@haswell$ N=6; for i in {1..10}; do perf stat same-function $N 2>&1 | grep GHz; done
6,234,770,361 cycles # 3.391 GHz
6,199,096,296 cycles # 3.391 GHz
6,213,348,126 cycles # 3.391 GHz
6,217,971,263 cycles # 3.391 GHz
6,224,779,686 cycles # 3.391 GHz
6,194,117,897 cycles # 3.391 GHz
6,225,259,274 cycles # 3.391 GHz
6,244,391,509 cycles # 3.391 GHz
6,189,972,381 cycles # 3.391 GHz
6,205,556,306 cycles # 3.391 GHz
考虑到这一点,我们重新阅读了所有英特尔架构手册中的每一个字,筛选整个网络上的每个页面,提到“计算机”这些词语。或编程&#39;并在山顶上孤独地冥想6年。没有实现任何启示,我们来到文明,刮胡子,洗个澡,并问StackOverflow的专家:
这里可能发生什么?
编辑:在本杰明的帮助下(请参阅下面的答案)我已经提出了更多succinct test case。它是一个独立的20行装配线。即使结果保持不变并且执行相同数量的指令,从使用SUB变为SBB也会导致性能差异达到15%。解释吗?我想我越来越接近一个了。
; Minimal example, see also http://stackoverflow.com/q/26266953/3766665
; To build (Linux):
; nasm -felf64 func.asm
; ld func.o
; Then run:
; perf stat -r10 ./a.out
; On Haswell and Sandy Bridge, observed runtime varies
; ~15% depending on whether sub or sbb is used in the loop
section .text
global _start
_start:
push qword 0h ; put counter variable on stack
jmp loop ; jump to function
align 64 ; function alignment.
loop:
mov rcx, 1000000000
align 64 ; loop alignment.
l:
mov rax, [rsp]
add rax, 1h
mov [rsp], rax
; sbb rcx, 1h ; which is faster: sbb or sub?
sub rcx, 1h ; switch, time it, and find out
jne l ; (rot13 spoiler: foo vf snfgre ol 15%)
fin: ; If that was too easy, explain why.
mov eax, 60
xor edi, edi ; End of program. Exit with code 0
syscall
答案 0 :(得分:9)
查看完整的perf stat输出,您将看到它不是指令的数量,而是停止的周期数。
看看反汇编,我发现了两件事:
在我的机器上进行上述更改测试然后产生:
for f in $( seq 7); do perf stat -e cycles -r3 ./same-function $f 2>&1; done|grep cycles
6,070,933,420 cycles ( +- 0.11% )
6,052,771,142 cycles ( +- 0.06% )
6,099,676,333 cycles ( +- 0.07% )
6,092,962,697 cycles ( +- 0.16% )
6,151,861,993 cycles ( +- 0.69% )
6,074,323,033 cycles ( +- 0.36% )
6,174,434,653 cycles ( +- 0.65% )
修改强> 我在每个函数中都使用了一个易失性成员,在我的I7-3537U上测试了不同的编译,发现`-falign-loops = 64&#39;实际上是最慢的:
$ gcc -std=gnu99 -O3 -Wall -Wextra same-function.c -o same-function
$ gcc -falign-loops=64 -std=gnu99 -O3 -Wall -Wextra same-function.c -o same-function-l64
$ gcc -falign-functions=64 -std=gnu99 -O3 -Wall -Wextra same-function.c -o same-function-f64
$ for prog in same-function{,-l64,-f64}; do echo $prog; for f in $(seq 7); do perf stat -e cycles -r10 ./$prog $f 2>&1; done|grep cycl; done
same-function
6,079,966,292 cycles ( +- 0.19% )
7,419,053,569 cycles ( +- 0.07% )
6,136,061,105 cycles ( +- 0.27% )
7,282,434,896 cycles ( +- 0.74% )
6,104,866,406 cycles ( +- 0.16% )
7,342,985,942 cycles ( +- 0.52% )
6,208,373,040 cycles ( +- 0.50% )
same-function-l64
7,336,838,175 cycles ( +- 0.46% )
7,358,913,923 cycles ( +- 0.52% )
7,412,570,515 cycles ( +- 0.38% )
7,435,048,756 cycles ( +- 0.10% )
7,404,834,458 cycles ( +- 0.34% )
7,291,095,582 cycles ( +- 0.99% )
7,312,052,598 cycles ( +- 0.95% )
same-function-f64
6,103,059,996 cycles ( +- 0.12% )
6,116,601,533 cycles ( +- 0.29% )
6,120,841,824 cycles ( +- 0.18% )
6,114,278,098 cycles ( +- 0.09% )
6,105,938,586 cycles ( +- 0.14% )
6,101,672,717 cycles ( +- 0.19% )
6,121,339,944 cycles ( +- 0.11% )
有关对齐循环与对齐函数的更多详细信息:
$ for prog in same-function{-l64,-f64}; do sudo perf stat -d -r10 ./$prog 0; done
Performance counter stats for './same-function-l64 0' (10 runs):
2396.608194 task-clock:HG (msec) # 1.001 CPUs utilized ( +- 0.64% )
56 context-switches:HG # 0.024 K/sec ( +- 5.51% )
1 cpu-migrations:HG # 0.000 K/sec ( +- 74.78% )
46 page-faults:HG # 0.019 K/sec ( +- 0.63% )
7,331,450,530 cycles:HG # 3.059 GHz ( +- 0.51% ) [85.68%]
5,332,248,218 stalled-cycles-frontend:HG # 72.73% frontend cycles idle ( +- 0.71% ) [71.42%]
<not supported> stalled-cycles-backend:HG
5,000,800,933 instructions:HG # 0.68 insns per cycle
# 1.07 stalled cycles per insn ( +- 0.04% ) [85.73%]
1,000,446,303 branches:HG # 417.443 M/sec ( +- 0.04% ) [85.75%]
8,461 branch-misses:HG # 0.00% of all branches ( +- 6.05% ) [85.76%]
<not supported> L1-dcache-loads:HG
45,593 L1-dcache-load-misses:HG # 0.00% of all L1-dcache hits ( +- 3.61% ) [85.77%]
6,148 LLC-loads:HG # 0.003 M/sec ( +- 8.80% ) [71.36%]
<not supported> LLC-load-misses:HG
2.394456699 seconds time elapsed ( +- 0.64% )
Performance counter stats for './same-function-f64 0' (10 runs):
1998.936383 task-clock:HG (msec) # 1.001 CPUs utilized ( +- 0.61% )
60 context-switches:HG # 0.030 K/sec ( +- 17.77% )
1 cpu-migrations:HG # 0.001 K/sec ( +- 47.86% )
46 page-faults:HG # 0.023 K/sec ( +- 0.68% )
6,107,877,836 cycles:HG # 3.056 GHz ( +- 0.34% ) [85.63%]
4,112,602,649 stalled-cycles-frontend:HG # 67.33% frontend cycles idle ( +- 0.52% ) [71.41%]
<not supported> stalled-cycles-backend:HG
5,000,910,172 instructions:HG # 0.82 insns per cycle
# 0.82 stalled cycles per insn ( +- 0.01% ) [85.72%]
1,000,423,026 branches:HG # 500.478 M/sec ( +- 0.02% ) [85.77%]
10,660 branch-misses:HG # 0.00% of all branches ( +- 13.23% ) [85.80%]
<not supported> L1-dcache-loads:HG
47,492 L1-dcache-load-misses:HG # 0.00% of all L1-dcache hits ( +- 14.82% ) [85.80%]
11,719 LLC-loads:HG # 0.006 M/sec ( +- 42.44% ) [71.28%]
<not supported> LLC-load-misses:HG
1.997319759 seconds time elapsed ( +- 0.62% )
这两个可执行文件的指令/周期数都非常低,可能是由于循环的简约性和它造成的内存压力,但我不知道为什么一个比另一个更差。
我也尝试了一些
的内容$ for prog in same-function{-l64,-f64}; do sudo perf stat -eL1-{d,i}cache-load-misses,L1-dcache-store-misses,cs,cycles,instructions -r10 ./$prog 0; done
和
$ sudo perf record -F25000 -e'{cycles:pp,stalled-cycles-frontend}' ./same-function-l64 0
[ perf record: Woken up 28 times to write data ]
[ perf record: Captured and wrote 6.771 MB perf.data (~295841 samples) ]
$ sudo perf report --group -Sloop0 -n --show-total-period --stdio
$ sudo perf annotate --group -sloop0 --stdio
但是找不到罪魁祸首没有任何成功。尽管如此,我觉得无论如何都要在这里记下它可能会有所帮助......
编辑2: 这是我对same-function.c的补丁:
$ git diff -u -U0
diff --git a/same-function.c b/same-function.c
index f78449e..78a5772 100644
--- a/same-function.c
+++ b/same-function.c
@@ -20 +20 @@ done
-volatile uint64_t counter = 0;
+//volatile uint64_t counter = 0;
@@ -22,0 +23 @@ COMPILER_NO_INLINE void loop0(void) {
+volatile uint64_t counter = 0;
@@ -31,0 +33 @@ COMPILER_NO_INLINE void loop1(void) {
+volatile uint64_t counter = 0;
@@ -40,0 +43 @@ COMPILER_NO_INLINE void loop2(void) {
+volatile uint64_t counter = 0;
@@ -49,0 +53 @@ COMPILER_NO_INLINE void loop3(void) {
+volatile uint64_t counter = 0;
@@ -58,0 +63 @@ COMPILER_NO_INLINE void loop4(void) {
+volatile uint64_t counter = 0;
@@ -67,0 +73 @@ COMPILER_NO_INLINE void loop5(void) {
+volatile uint64_t counter = 0;
@@ -76,0 +83 @@ COMPILER_NO_INLINE void loop6(void) {
+volatile uint64_t counter = 0;
@@ -85,0 +93 @@ COMPILER_NO_INLINE void loop7(void) {
+volatile uint64_t counter = 0;
@@ -94,0 +103 @@ COMPILER_NO_INLINE void loop8(void) {
+volatile uint64_t counter = 0;
@@ -103,0 +113 @@ COMPILER_NO_INLINE void loop9(void) {
+volatile uint64_t counter = 0;
@@ -135 +145 @@ int main(int argc, char** argv) {
-}
\ No newline at end of file
+}
编辑3 :更简单的例子也是如此:
; Minimal example, see also http://stackoverflow.com/q/26266953/3766665
; To build (Linux):
; nasm -felf64 func.asm
; ld func.o
; Then run:
; perf stat -r10 ./a.out
; Runtime varies ~10% depending on whether
section .text
global _start
_start:
push qword 0h ; put counter variable on stack
jmp loop ; jump to function
;align 64 ; function alignment. Try commenting this
loop:
mov rcx, 1000000000
;align 64 ; loop alignment. Try commenting this
l:
mov rax, [rsp]
add rax, 1h
mov [rsp], rax
sub rcx, 1h
jne l
fin: ; End of program. Exit with code 0
mov eax, 60
xor edi, edi
syscall
这里效果相同。有趣。
干杯, 本杰明
答案 1 :(得分:1)
几年前,我会告诉你检查CPU到达任何一个循环时内部状态的差异;因为众所周知,这会对无序预测过程(或类似的东西)的容量产生深远的影响。例如,同一个循环的性能可能会改变15-20%,具体取决于CPU在进入循环之前所做的事情,只是从两个不同点跳转的事实可能足以改变执行速度。
在你的情况下,这是非常值得测试的。您所要做的就是改变IF块中指令的顺序;例如替换以下内容:
switch (firstLetter) {
case '0': loop0(); break;
case '1': loop1(); break;
case '2': loop2(); break;
case '3': loop3(); break;
case '4': loop4(); break;
case '5': loop5(); break;
case '6': loop6(); break;
case '7': loop7(); break;
case '8': loop8(); break;
case '9': loop9(); break;
default: goto die_usage;
}
使用:
switch (firstLetter) {
case '9': loop9(); break;
case '8': loop8(); break;
case '7': loop7(); break;
case '6': loop6(); break;
case '5': loop5(); break;
case '4': loop4(); break;
case '3': loop3(); break;
case '2': loop2(); break;
case '1': loop1(); break;
case '0': loop0(); break;
default: goto die_usage;
}
或任何随机订单。当然,您应该检查生成的汇编代码,以确保编译器没有对指令的顺序进行重新排序。
另外,因为你的循环在个别函数内部;你还应该确保这些函数本身在64字节边界上对齐。